On some generalizations of abelian power avoidability
On some generalizations of abelian power avoidability Michaël Rao LIP, CNRS, ENS de Lyon, UCBL, Université de Lyon, France
Abstract We prove that 2-abelian-cubes are avoidable over a binary alphabet and that 3-abelian-squares are avoidable over a ternary alphabet, answering positively to two questions of Karhumäki et al.. We also show the existence of infinite additive-cube-free words on several ternary alphabets. To achieve this, we give sufficient conditions for a morphism to be k-abelian-n-power-free (resp. additive-n-power-free), and then we give several morphisms which respect these conditions. Additionally, all our constructions show that the number of such words grows exponentially. As a corollary, we get a new lower bound of 31/19 = 1.059526 . . . for the growth rate of abelian-cube-free words. Keywords: Combinatorics on words, k-abelian equivalence, square-free, cube-free, morphism
1. Introduction Avoidability of repetitions in words is one of the most studied topics in word combinatorics since the seminal papers of Thue [26, 27]. One famous example is Dejean’s conjecture, recently solved by several authors (see [23]). The avoidability of abelian repetitions received a lot of interest since a question from Erdös in 1957 [9, 10]. Two words u, v ∈ A∗ are abelian equivalent, denoted u ≡a v, if for every a ∈ A, |u|a = |v|a . A word u is an abelian-n-power , where n ≥ 2, if u = u1 u2 . . . un such that ui ≡a ui+1 for every i ∈ {1, . . . , n − 1}. An abelian square (resp. abelian cube) is an abelian-2-power (resp. abelian-3-power). It is not difficult to see that every ternary word of size at least 8 has an abelian square. Erdös [9, 10] raised the question whether they can be avoided in an infinite word on an alphabet of size 4. Evdokimov [11] showed that one can avoid them on an alphabet of size 25, which was later lowered to 5 by Pleasants [22]. Finally, Keränen [18] answered positively to Erdos’s question in 1992. Furthermore, Dekking [7] showed that abelian cubes can be avoided in an infinite ternary word, and that abelian-4-powers can be avoided in an infinite binary word.
Email address: [email protected] (Michaël Rao)
Preprint submitted to Elsevier
December 15, 2014
We are here interested in two variations of the previous problem. The first one is the k-abelian-equivalence introduced by Karhumäki et al. [14, 16, 17]. Let k ≥ 1. Two words u and v (u, v ∈ A∗ ) are k-abelian-equivalent, denoted u ≡a,k v, if for every w ∈ A∗ with |w| ≤ k, |u|w = |v|w . A word u is a k-abeliann-power , n ≥ 2, if u = u1 u2 . . . un such that ui ≡a,k ui+1 for every i ∈ {1, . . . , n − 1}. A k-abelian-square (resp. k-abelian-cube) is a k-abelian-2-power (resp. k-abelian-3-power). This notion is between the abelian equivalence (which is the 1-abelian-equivalence) and the usual equality between words (which can be viewed as the ∞-abelian-equivalence). Since cubes are avoidable in the binary alphabet (e.g. in the Prouhet-Thue-Morse word), but are not avoidable in the abelian sense, it is natural to ask for the smallest k for which k-abelian-cubes are avoidable on a binary alphabet. In [14] authors showed that k ≤ 8, and in [20] that k ≤ 5. Finally, in [21], Mercaş and Saarela showed that k ≤ 3. The same question can be asked for k-abelian-squares on a ternary alphabet: 2-abeliansquares cannot be avoided [15], but Huova showed that 64-abelian-squares can be avoided [12]. In Section 2, we give sufficient conditions for a morphism h : A∗ → B ∗ to be k-abelian-n-power-free (for a fixed n ≥ 2 and k ≥ 1), that is for every abelian-n-power-free word w ∈ A∗ , h(w) is k-abelian-n-power-free. Then we give morphisms which respect the conditions, in order to construct 2-abeliancube-free binary words and 3-abelian-square-free ternary words. This answers the two previous questions and also prove that the number of such words grows exponentially, as abelian-square-free on four letters [3], and abelian-cube-free ternary words ([1], see also Section 3). The second notion is the additive-cube-avoidability. A word w ∈ N∗ is an additive cube ifP w = pqr, p, q and r are non-empty-word such that P whereP |p| = |q| = |r| and (p) = (q) = (r). A word is additive-cube-free if it has no factor which is an additive cube. Clearly, such words are also abelian-cubefree. Recently Cassaigne et al. [5] showed that one can construct an infinite additive-cube-free word on the alphabet {0, 1, 3, 4}. The question of infinite additive-square-free word’s existence on a finite alphabet is still open. ∗ In Section 3 we give sufficient conditions for a substitution h : A∗ → 2B , A, B ⊆ N, to be additive-cube-free. We present substitutions from the alphabet {0, 1, 3, 4} to several ternary alphabets which respects these conditions. Moreover, the presented constructions show directly that the number of additive-cubefree words on these ternary alphabets grows exponentially. The lower bound of 31/19 = 1.059526 . . . we obtain for the growth rate for the alphabet {0, 1, 8} is also a new lower bound for the number of abelian-cube-free words on a ternary alphabet. 2. k-abelian-n-power-free morphisms 2.1. Preliminaries Let |u|w denote the number of occurrences of the factor w in u. The Parikh vector of a word u ∈ A∗ , where A = {a1 , a2 , . . . , ak }, is Ψ(u) = (|u|a1 , |u|a2 , . . . , |u|ak ). For a set S ⊆ A∗ , ΨS (u) is the vector indexed by S such that 2
ΨS (u)[w] = |u|w for every w ∈ S. When the alphabet is clear in the context, we let Ψk (u) be ΨAk (u), for k ≥ 1. Let Pref(u) be the set of prefixes of u, and Suf(u) be its set of suffixes. For k ≥ 0, let pref k (u) (resp. suf k (u)) be the prefix (resp. suffix) of u of size k. There are several equivalent definitions for k-abelian-equivalence (see [17]). Two words u and v of size at most k − 1 are k-abelian-equivalent if and only if they are equal. Otherwise, the following conditions are equivalent: • u and v are k-abelian-equivalent (i.e. u ≡a,k v). • For every w ∈ A∗ with |w| ≤ k, |u|w = |v|w . • For every w ∈ Ak , |u|w = |v|w , pref k−1 (u) = pref k−1 (v) and suf k−1 (u) = suf k−1 (v). • For every w ∈ Ak , |u|w = |v|w , and pref k−1 (u) = pref k−1 (v). Given k ≥ 1 and n ≥ 2, a (possibly infinite) word w is k-abelian-n-power-free if no non-empty factor in w is a k-abelian-n-power. A word is k-abelian-squarefree (resp. k-abelian-cube-free) if it is k-abelian-2-power-free (resp. k-abelian-3power-free). A morphism h : A∗ → B ∗ is k-abelian-n-power-free if for every abelian-npower-free word u ∈ A∗ , h(u) is k-abelian-n-power-free. Note that u has to be abelian-n-power-free, not only k-abelian-n-power-free; we explain in Section 2.4 why we use this weaker notion. A morphism h : A∗ → B ∗ is k-abelian-squarefree (resp. k-abelian-cube-free) if it is k-abelian-2-power-free (resp. k-abelian-3power-free). 2.2. Testing k-abelian-n-power-freeness In [2], Carpi gave a set of conditions which assures that a given morphism is abelian-n-power-free. We give in the following theorem a set of similar conditions which assures that a given morphism is k-abelian-n-power-free. Theorem 1. We fix k ≥ 1 and n ≥ 2, and two alphabets A and B. Let h : A∗ → B ∗ be a morphism. Suppose that: (i) For every abelian-n-power-free word w ∈ A∗ with |w| ≤ 2 or |h(w[2 : |w| − 1])| ≤ (k − 2)n − 2, h(w) is k-abelian-n-power-free. (ii) There are p, s ∈ B k−1 such that for every a ∈ A, p = pref k−1 (h(a)p) and s = suf k−1 (sh(a)). (iii) The matrix N indexed by B k × A, with N [w, x] = |h(x)p|w , has rank |A|. (iv) Let S ⊆ B k , with |S| = |A|, such that the matrix M indexed by S × A, with M [w, x] = |h(x)p|w , is invertible. Let ΨS (v, u) = ΨS (vp) + ΨS (su) − ΨS (sp) and Ψk (v, u) = ΨB k (v, u). For every ai ∈ A and ui , vi ∈ A∗ with ui vi = h(ai ); 0 ≤ i ≤ n; such that: 3
(P) |{pref k−1 (vi p) : 0 ≤ i < n}| = 1, (I) M −1 (ΨS (vi−1 , ui ) − ΨS (vi , ui+1 )) is an integer vector, for every 1 ≤ i < n, (C) Ψk (vi−1 , ui ) − Ψk (vi , ui+1 ) ∈ im(N ) for every 1 ≤ i < n, there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that for every 1 ≤ i < n : M −1 ΨS (vi−1 , ui ) − (1 − αi−1 )Ψ(ai−1 ) − αi Ψ(ai ) = M −1 ΨS (vi , ui+1 ) − (1 − αi )Ψ(ai ) − αi+1 Ψ(ai+1 ). (1) Then h is k-abelian-n-power-free. Proof. Suppose that h(w) has a k-abelian-n-power q1 . . . qn . Let q0 and qn+1 be such that h(w) = q0 q1 . . . qn qn+1 . By condition (i), if |q1 | < k − 1, then w has an abelian-n-power. So we have |qi | ≥ k − 1 for every 1 ≤ i ≤ n. There are, for every 0 ≤ i ≤ n, ai ∈ A, ui ∈ Pref(h(ai )) and ri ∈ A∗ such that, for every 0 ≤ i ≤ n, r0 . . . ri ai ∈ Pref(w) and q0 . . . qi = h(r0 . . . ri )ui . Note that, for a 1 ≤ i ≤ n, ri can be empty, but ai is always the first letter of ri+1 ai+1 . Let vi be such that ui vi = h(ai ) for every 0 ≤ i ≤ n. By condition (i), one can suppose w.l.o.g. that |r1 . . . rn an | ≥ 3. By condition (ii), for every 1 ≤ i ≤ n, pref k−1 (qi ) = pref k−1 (vi−1 p). Since q1 . . . qn is a k-abelian-n-power, we have condition (P). Claim 1. Let r ∈ A∗ and u, v ∈ B ∗ . Then: • N Ψ(r) = Ψk (h(r)p) = Ψk (sh(r)) = Ψk (sh(r)p) − Ψk (sp), • Ψk (vh(r)p) = Ψk (vp) + N Ψ(r), • Ψk (sh(r)u) = Ψk (su) + N Ψ(r). Proof. If pref k−1 (u) = p, then Ψk (vu) = Ψk (vp) + Ψk (u). Similarly, if suf k−1 (v) = s, then Ψk (vu) = Ψk (v) + Ψk (su). All the equalities follow from the previous facts, and the definition of N . Claim 2. For every 1 ≤ i ≤ n: Ψk (qi ) = N (Ψ(ri ) − Ψ(ai−1 )) + Ψk (vi−1 , ui ).
(2)
Proof. By double counting, we have : Ψk (qi ) + Ψk (sh(ri ai )p) = Ψk (sh(ri )ui ) + Ψk (vi−1 h(a−1 i−1 ri ai )p). By Claim 1: Ψk (qi ) + N Ψ(ri ai ) + Ψk (sp) = Ψk (sui ) + N Ψ(ri ) + Ψk (vi−1 p) + N Ψ(a−1 i−1 ri ai ). Thus: Ψk (qi ) = Ψk (vi−1 , ui ) + N (Ψ(ri ) − Ψ(ai−1 )). 4
Since Ψk (qi ) = Ψk (qi+1 ) for every 1 ≤ i < n, we have the condition (C). Now we have directly ΨS (qi ) = M (Ψ(ri ) − Ψ(ai−1 )) + ΨS (vi−1 , ui ). Since ΨS (qi ) = ΨS (qi+1 ): M −1 (ΨS (vi−1 , ui ) − ΨS (vi , ui+1 )) = Ψ(ri+1 ) − Ψ(ai ) − Ψ(ri ) + Ψ(ai−1 ). The right part is an integer vector, so we have condition (I). Thus, by condition (iv), there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that (1) is fulfilled. Equation (1) together with equation (2) give: − Ψ(ri ) + Ψ(ai−1 ) − (1 − αi−1 )Ψ(ai−1 ) − αi Ψ(ai ) = −Ψ(ri+1 ) + Ψ(ai ) − (1 − αi )Ψ(ai ) − αi+1 Ψ(ai+1 ) that is: Ψ(ri ) − αi−1 Ψ(ai−1 ) + αi Ψ(ai ) = Ψ(ri+1 ) − αi Ψ(ai ) + αi+1 Ψ(ai+1 ).
(3)
In equation (3), either the left or the right part is a non-negative vector. Since equation (3) is fulfilled for every 1 ≤ i < n, Ψ(ri ) − αi−1 Ψ(ai−1 ) + αi Ψ(ai ) is −α i a non negative vector for every 1 ≤ i ≤ n. Let ri0 = ai−1i−1 ri aα i ; 1 ≤ i ≤ n. 0 Since ai is the first letter of ri ai+1 , and Ψ(ri ) = Ψ(ri ) − αi−1 Ψ(ai−1 ) + αi Ψ(ai ) is a non-negative vector, ri0 is well defined in B ∗ . In one hand r10 . . . rn0 is a factor of w, and is non empty since |r10 . . . rn0 | ≥ |r1 . . . rn an | − 2. On the other 0 hand Ψ(ri0 ) = Ψ(ri+1 ) (by equation 3), for every 1 ≤ i < n. Thus, w has an abelian-n-power r10 . . . rn0 . We introduce ΨS (v, u) in order to handle pairs (v, u) such that |vu| < k − 1 (otherwise we have Ψk (v, u) = Ψk (vu)). Theorem 1 gives a set of sufficient conditions, but are still far from a characterization, as Carpi partially done for abelian-n-power-free morphisms [2]. The key point is the condition (ii). One mentions that we can save up the suffix condition in (ii) by carefully handling the cases where ui or vi has size less than k. However, we still need either the prefix (or the suffix) condition in order to properly define N . 2.3. 2-abelian-cube-free and 3-abelian-square-free morphisms Morphisms h2 and h02 respect the conditions of Theorem 1 for k = 2 and n = 3, i.e. are 2-abelian-cube-free, while morphisms h3 and h03 respect the conditions for k = 3 and n = 2, i.e. are 3-abelian-square-free. The checks were done by computer, and took only a few seconds. Thus, the infinite word h2 (u) (resp. h02 (u)) where u is an infinite abelian-cube-free word (for example a fixed point of Dekking’s morphism µ : 0 → 0012, 1 → 112, 2 → 022 [7]) is a 2-abelian-cube-free binary word. Similarly, h3 (v) (resp. h03 (v)), where v is an infinite abelian-square-free word on an alphabet of size 4 (for example, a fixed point of Keränen’s morphism g85 [18]), is an infinite 3-abelian-square-free ternary word. Over all the 2-abelian-cube-free morphisms we found, h2 is the smallest uniform morphism, while h02 is the one which minimize |h(012)|. If we are only 5
2-abelian-cube-free morphisms: 0 → 00100101001011001001010010011001001100101101011 h2 : 1 → 00100110010011001101100110110010011001101101011 2 → 00110110101101001011010110100101001001101101011 0 → 00100101001100100101001001100100110011011 0 h2 : 1 → 010110110011011001100100110011011 2 → 0101101001010010110011011 3-abelian-square-free morphisms: 0 → 0102012021012010201210212 1 → 0102101201021201210120212 h3 : 2 → 0102101210212021020120212 3 → 0121020120210201210120212 0 → 01201020120212012101201021 1 → 01202120121021201021 h03 : 2 → 0120210201021 3 → 0121020121 Morphisms such that h(µ∞ (0)) is 2-abelian-cube-free: 0 → 001001100110110011001001100100101 hd : 1 → 001011010110100101001001100100101 2 → 001011010110110011001001101011011 0 → 0101101001011 0 hd : 1 → 010110110011011001100100110011011 2 → 00100101001001100100110011011 Table 1: Morphisms for k-abelian-n-power-free words.
6
interested in 2-abelian-cube-free infinite word, one can find simpler construction. The morphism hd ◦ µ is 2-abelian-cube-free so hd (µ∞ (0)) is 2-abelian-cube-free. We also claim that h0d (µ∞ (0)) is 2-abelian-cube-free. One can modify the decision procedure of Theorem 1 to compute the set of “patterns” that u has to avoid to ensure that h(u) is k-abelian-n-power-free. This notion of patterns was used by Carpi [3, 4] to prove that a substitution is abelian-square free, or by Keränen [19] to prove that a fixed point of g98 is abelian-square free, even though g98 is not abelian-square free. This was also used, under the name of template, by Aberkane et al. [1] to show the exponential growth rate of abelian-cube-free ternary words, and by Currie and Rampersad [6] for an algorithm which decide if a fixed point of a morphism is abelian-n-power-free. More recently, Mercaş and Saarela [20, 21] used this kind of patterns to show that a morphic word is k-abelian-cube-free. Doing this, we are able to show that h0d ◦ µ3 (u) is 2-abelian-cube free if and only if u forbids factors of the form F = {pqr, 1p0q0r2 : Ψ(p) = Ψ(q) = Ψ(r)} ∪ {0p1q0r2, 1p1q0r2 : Ψ(p1) = Ψ(q0) = Ψ(r0)}. Moreover, µ(u) forbids factors of the form F if and only if u forbids factors of the form F (in other words, µ is F -free). Thus, h0d (µ∞ (0)) is 2-abelian-cube-free, but for every n ≥ 0, h0d ◦ µn is not 2-abelian-cube-free (e.g. for every n ≥ 0, h0d (µn (1002)) has a 2abelian-cube). 2.4. Final remarks and questions We finally shortly explain why we use this weak notion of k-abelian-n-powerfreeness for morphisms. On one hand, k-abelian-squares cannot be avoided by a pure morphic word on a ternary alphabet [13]. So there is no morphism h : {0, 1, 2} → {0, 1, 2} such that for every k-abelian-square-free word u, h(u) is k-abelian-square-free, except trivial ones. On the other hand, suppose that there is a morphism h : A∗ → B ∗ , with |A| > |B|, such that for every 2-abeliancube-free word u ∈ A∗ , h(u) is 2-abelian-cube-free. Without lost of generality, there is {a, b} ⊆ A, such that the first letter of h(a) and h(b) is the same. Then babbababb is 2-abelian-cube-free, but h(bab) ≡a,2 h(abb) thus h(babbababb) is an 2-abelian-cube. We have a contradiction, so such a morphism cannot exist. Nevertheless, we cannot conclude directly when |A| = |B| and the first and last letters of the images differ. More specifically, the following question is still open. Question 1. Is there a pure morphic binary word which avoids 2-abelian-cubes ? Let us also raise some questions on avoidability of long repetitions. Every infinite binary word contains arbitrarily long abelian squares, while ones exist which avoid squares of period at least 3 [8, 25]. We recently showed that one can avoid 3-abelian-squares of period at least 3 over a binary alphabet [24]. It seems natural to ask the following: Question 2. Is there a p ∈ N such that 2-abelian-squares of period at least p can be avoided over a binary alphabet ? 7
That reminds the questions suggested by Mäkelä (see [19]): Question 3. (1) Can we avoid abelian-squares of the form uv, with |u| ≥ 2, over a ternary alphabet ? (2) Can we avoid abelian-cubes of the form uvw, with |u| ≥ 2, over a binary alphabet ? In [24], we answer negatively to (1). Then we modify this question to the following one: Question 4. Is there a p ∈ N such that one can avoid abelian cubes of period at least p over a binary alphabet ? 3. Ternary words avoiding additive cubes 3.1. Testing additive-n-power-freeness Let Σ be the morphism from the free monoid on the alphabet N to the additive group (Z, +) such that Σ(x) = x for every x ∈ N. A word w ∈ N∗ is an additive-n-power , with n ≥ 2, if w = p1 . . . pn , such that for every 1 ≤ i < n, |pi | = |pi+1 | and Σ(pi ) = Σ(pi+1 ). A word is an additive-cube (resp. additivesquare) if it is an additive-3-power (additive-2-power). A (possibly infinite) word w is additive-n-power-free if no non-empty factor of w is an additive-n-power. Clearly, such words are also abelian-n-power-free. In [5], authors prove that the fixed point of the morphism 0 → 03, 1 → 43, 3 → 1, 4 → 01 is additive-cube-free. ∗ A substitution is a morphism s : A∗ → 2B between the free monoid A∗ and the power monoid of B ∗ , that is the monoid of subsets of B ∗ , with the operation U · V = {uv : (u, v) ∈ U × V }. A morphism h : A∗ → B ∗ can be ∗ viewed as a substitution s : A∗ → 2B such that s(w) = {h(w)}. A substitution ∗ s : A∗ → 2B , where A, B ⊆ N, is additive-n-power-free if for every additive-npower-free word u ∈ A∗ , every v ∈ s(u) is additive-n-power-free. We give sufficient conditions for a substitution to be additive-n-power-free in the following theorem. ∗
Theorem 2. We fix n ≥ 2 and A, B ⊆ N. Let s : A∗ → 2B be a substitution. Suppose that: (i) For every additive-n-power-free word w0 ∈ A∗ with |w0 | ≤ 2, every w ∈ s(w0 ) is additive-n-power-free. (ii) There is (l, γ, β) ∈ N × Z × Z, with β 6= 0, such that for every a ∈ A and w ∈ s(a), we have |w| = l and Σ(w) = γ + aβ. (iii) For every ai ∈ A, wi ∈ s(ai ), and ui , vi ∈ A∗ with ui vi = wi ; 0 ≤ i ≤ n; such that for every 1 ≤ i < n: (L) |vi−1 ui | ≡ |vi ui+1 | (mod l), 8
s015
s016
s017
s027
s037
s018
s038
s019
s029
s049
0 → {005015100100115010115, 005015100100115100115} 1 → {005015100100105055115, 050015100100105055115} : 3 → {005015101155155055115, 050015101155155055115} 4 → {005015155055155055115, 050015155055155055115} 0 → {00101160101006001016, 00101160101006001106} 1 → {00166060101006001016, 00166060101006001106} : 3 → {00166166110160661106, 00166166110166061106} 4 → {00166166066160661106, 00166166066166061106} 0 → {00170010011711001071, 00170010011711001701} 1 → {00170017707001001071, 00170017707001001701} : 3 → {00170017711017177077, 01070017711017177077} 4 → {00170017707077177077, 01070017707077177077} 0 → {0020720220220722007, 0020720220227022007} 1 → {7220720220220722007, 7220720220227022007} : 3 → {7077200770720722007, 7077200770727022007} 4 → {7077272770720722007, 7077272770727022007} 0 → {00300307303037707307, 00300307303037700737, 00300307303037707037} 1 → {00300300707737700737, 00300300707737707037, 00300300707737707307} : 3 → {00337730337737700737, 00337730337737707037, 00337730337737707307} 4 → {00337737707737700737, 00337737707737707307, 00337737707737707037} 0 → {0081001008011811011, 0081010080011811011, 0081001080011811011} 1 → {0081001008011818008, 0081010080011818008, 0081001080011818008} : 3 → {0081018818808811811, 0081108818808811811, 0081810818808811811} 4 → {0081018818808808188, 0081108818808808188, 0081188018808808188} 0 → {003800303830033833003, 003800308330033833003} 1 → {003800303830080038388, 003800308330080083838} : 3 → {003808833833038838838, 003808838330338838838} 4 → {003808838388088388388, 083008838388088388388} 0 → {0090110191001009, 0090110911001009} 1 → {0090119110110199, 0900119110110199} : 3 → {0090190090099199, 0900190090099199} 4 → {0090119199099199, 0900119199099199} 0 → {00290020020090022029, 00290020020090020229} 1 → {00290099220090022029, 00290099220090020229} : 3 → {00220292299099299099, 22920220099099299099} 4 → {22920992299099299099, 22990292299099299099} 0 → {00400400900499009, 00400400900949009} 1 → {00400449440099409, 00400449440499009} : 3 → {00409909909499409, 00409909949099409} 4 → {44944909949499009, 44944909949909409} Table 2: Additive-cube-free substitutions.
9
(M) Σ(vi−1 ui ) ≡ Σ(vi ui+1 ) + xi γ (mod β), (where xi = (|vi−1 ui | − |vi ui+1 |)/l for every 1 ≤ i < n) there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that for every 1 ≤ i < n: (a) αi − αi−1 = xi + αi+1 − αi , (b) Σ(vi−1 ui ) + β[(αi−1 − 1)ai−1 − αi ai ] = Σ(vi ui+1 ) + γxi + β[(αi − 1)ai − αi+1 ai+1 ]. Then s is additive-n-power-free. Proof. Suppose that w ∈ s(w0 ) has an additive-n-power q1 . . . qn . Let q0 and qn+1 be such that w = q0 q1 . . . qn qn+1 . For every 0 ≤ i ≤ n, there is ai ∈ A, wi ∈ s(ai ), ui ∈ Pref(wi ), and ri ∈ A∗ such that r0 . . . ri ai ∈ Pref(w0 ) and q0 . . . qi ∈ s(r0 . . . ri ) · {ui }. Let vi be such that ui vi = wi for every 0 ≤ i ≤ n. By condition (i), one can suppose w.l.o.g. that |r1 . . . rn an | ≥ 3. By condition (ii), for every p ∈ s(p0 ), we have Σ(p) = γ|p0 | + βΣ(p0 ). For every 1 ≤ i ≤ n, we have ui−1 qi ∈ s(ri ) · {ui }. Thus, by condition (ii), and by the fact that ui−1 vi−1 = wi−1 we have: |qi | = |vi−1 ui | + l(|ri | − 1)
(4)
Σ(qi ) = γ(|ri | − 1) + β(Σ(ri ) − ai−1 ) + Σ(vi−1 ui ).
(5)
and By equation (4) and by the fact that for every 1 ≤ i < n, |qi | = |qi−1 |, we have the condition (L), and: |ri+1 | − |ri | = (|vi−1 ui | − |vi ui+1 |)/l = xi . Since for every 1 ≤ i < n, Σ(qi ) = Σ(qi−1 ), we have: γ(|ri | − 1) + β(Σ(ri ) − ai−1 ) + Σ(vi−1 ui ) = γ(|ri+1 | − 1) + β(Σ(ri+1 ) − ai ) + Σ(vi ui+1 ). (6) Thus Σ(vi−1 ui ) = Σ(vi ui+1 ) + γxi + β(Σ(ri+1 ) − ai − Σ(ri ) + ai−1 ), and equation (M) is fulfilled. So, by condition (iii), there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that (a) and (b) are fulfilled. By equation (a), we have, for every 1 ≤ i < n; |ri | + αi − αi−1 = |ri−1 | + αi+1 − αi .
(7)
If ri is empty, ai = ai+1 otherwise the first letter of ri is ai . In equation (7), the right side or the left side must be non-negative. Thus, for every 1 ≤ i ≤ n, 10
−α
i |ri | + αi − αi−1 ≥ 0, and ri0 = ai−1i−1 ri aα i ; 1 ≤ i ≤ n; is well defined. We have 0 0 |ri | = |ri | + αi − αi−1 and Σ(ri ) = Σ(ri ) + αi ai − αi−1 ai−1 . By equation (7), 0 for every 1 ≤ i < n, |ri0 | = |ri+1 |. Moreover r10 . . . rn0 is a factor of w0 , and is non 0 0 empty since |r1 . . . rn | ≥ |r1 . . . rn an | − 2. 0 ). Thus, Σ(ri0 ) = When we subtract (b) to (6), we get βΣ(ri0 ) = βΣ(ri+1 0 0 Σ(ri+1 ) for every 1 ≤ i < n, and w has an additive-n-power r10 . . . rn0 .
Theorem 2 can be used to find additive-square-free, additive-cube-free and additive-4-power-free substitutions. However, we have few hopes to find an additive-square-free substitution, while additive-4-powers are equivalent to abelian-4-powers on binary words. 3.2. Additive-cube-free substitutions We have checked by computer that every substitution in Table 2 respects the conditions of Theorem 2. Since there is an infinite additive-cube-free word on the alphabet {0, 1, 3, 4}, one can construct infinite additive-cube-free words on the alphabets {0, 1, 5}, {0, 1, 6}, {0, 1, 7}, {0, 2, 7}, {0, 3, 7}, {0, 1, 8}, {0, 3, 8}, {0, 1, 9}, {0, 2, 9} and {0, 4, 9}. In our substitutions, each letter has at least two images. This clearly shows that number of additive-cube-free words on these alphabets grows exponentially. For the alphabet {0, 1, 8}, we got 3 images of size 19 for each letter, giving the lower bound of 31/19 = 1.059526 . . . for the growth rate. This bound is also a new lower bound for the growth rate of abelian-cube-free words on ternary alphabet. (The previous known bound was 21/24 = 1.029302 . . . in [1].) We conjecture that for every alphabet A = {0, i, j} such that i and j are coprime and j ≥ 6, there exists an infinite additive-cube-free word on the alphabet A. The cases {0, 1, 2}, {0, 1, 3}, {0, 1, 4} and {0, 2, 5} are left open. Furthermore, it seems difficult to construct a very long word on the alphabet {0, 1, 2, 3} avoiding additive cubes (the longest we got has size ∼ 1.4 × 105 ). Question 5. Are there infinite additive-cube-free words on the following alphabets : {0, 1, 2, 3}, {0, 1, 4} and {0, 2, 5} ? The substitutions in Table 3 also respect the conditions of Theorem 2, thus the existence of an infinite additive-cube-free word on the alphabet {0, 1, 2, 3} imply the existence of infinite additive-cube-free words on alphabets {0, 1, 4} and {0, 2, 5}. Acknowledgments The author would like to thank Pascal Ochem for valuable discussions on these problems. [1] A. Aberkane, J. D. Currie, N. Rampersad. The number of ternary words avoiding abelian cubes grows exponentially. J. Integer Sequences, Vol. 7.2, Art. 04.2.7 (2004).
11
s014
s025
0 → {004114104011011004011} 1 → {004114104011011014144} : 2 → {004114104010044044144} 3 → {004114104044144044144} 0 → {02200520220250552} 1 → {02252520220250552} : 2 → {02255055200550552} 3 → {02255055252550552}
Table 3: Additive-cube-free substitutions from {0, 1, 2, 3}∗ .
[2] A. Carpi. On Abelian Power-Free Morphisms. Int. J. Algebra Comput., Vol. 03(2) 151–167 (1993). [3] A. Carpi. On the number of Abelian square-free words on four letters. Disc. Appl. Math, Vol. 81(1-3), 155–167 (1998). [4] A. Carpi. On Abelian squares and substitutions. Theoret. Comput. Sci., Vol. 218(1), 61–81 (1999). [5] J. Cassaigne, J. D. Currie, L. Schaeffer, J. Shallit. Avoiding three consecutive blocks of the same size and same sum. J. ACM, Vol. 61(2), Art. 10 (2014). [6] J. D. Currie, N. Rampersad. Fixed points avoiding Abelian k-powers. J. Comb. Theory Ser. A, Vol. 119(5) 942–948 (2012). [7] F.M. Dekking. Strongly nonrepetitive sequences and progression-free sets. J. Comb. Theory Ser. A, Vol. 27(2), 181–185 (1979). [8] R.C Entringer, D.E Jackson, J.A Schatz. On nonrepetitive sequences. J. Comb. Theory Ser. A, Vol. 16(2), 159–164 (1974). [9] P. Erdös. Some unsolved problems. Michigan Math. J., Vol. 4(3), 291–300 (1957). [10] P. Erdös. Some unsolved problems. Magyar Tud. Akad. Mat. Kutató Int. Közl, Vol. 6, 221–254 (1961). [11] A. A. Evdokimov. Strongly asymmetric sequences generated by a finite number of symbols. Dokl. Akad. Nauk. SSSR, Vol. 179, 1268–1271 (1968); Soviet Math. Dokl., Vol. 9, 536–539 (1968). [12] M. Huova. Existence of an infinite ternary 64-abelian square-free word. RAIRO - Theor. Inf. and Applic., Vol. 48(3), 307–314 (2014).
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[13] M. Huova, J. Karhumäki. On the Unavoidability of k-Abelian Squares in Pure Morphic Words. J. Integer Sequences, Vol. 16, Art. 13.2.9 (2013). [14] M. Huova, J. Karhumäki, A. Saarela. Problems in between words and abelian words: k-abelian avoidability. Theoret. Comput. Sci., Vol. 454, 172–177 (2012). [15] M. Huova, J. Karhumäki, A. Saarela, K. Saari. Local squares, periodicity and finite automata. Rainbow of Computer Science, LNCS Vol. 6570, 90– 101 (2011). [16] J. Karhumäki. Generalized Parikh mappings and homomorphisms. Inform. Control, Vol. 47, 155–165 (1980). [17] J. Karhumäki, A. Saarela, L. Zamboni. On a generalization of Abelian equivalence and complexity of infinite words. J. Comb. Theory Ser. A, Vol. 120(8), 2189–2206 (2013). [18] V. Keränen. Abelian squares are avoidable on 4 letters. Proc. of ICALP 1992, LNCS, Vol. 623, 41–52 (1992). [19] V. Keränen. New Abelian Square-Free DT0L-Languages over 4 Letters. Manuscript (2003). [20] R. Mercaş, A. Saarela. 5-abelian cubes are avoidable on binary alphabets. Proc. of the 14th Mons Days of Theoret. Comput. Sci. (2012). [21] R. Mercaş, A. Saarela. 3-Abelian Cubes Are Avoidable on Binary Alphabets. Proc. of DLT 2013, LNCS Vol. 7907, 374–383 (2013). [22] P. A. B. Pleasants. Non-repetitive sequences. Proc. Cambridge Philos. Soc., Vol. 68, 267–274 (1970). [23] M. Rao. Last cases of Dejean’s conjecture. Theoret. Comput. Sci., Vol. 412(27), 3010–3018 (2011). [24] M. Rao, M. Rosenfeld. Avoidability of long k-abelian repetitions. Proc. of the 15th Mons Days of Theoret. Comput. Sci. (2014). [25] A. S. Fraenkel, R. J. Simpson. How many squares must a binary sequence contain? Electron. J. Combin., Vol. 2, (1995). [26] A. Thue. Über unendliche Zeichenreihen. Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, Vol. 7, 1–22 (1906). [27] A. Thue. Über die gegenseitige Lage gleicher Teile gewisser Zeichenreihen. Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, Vol. 10, 1–67 (1912).
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Abstract We prove that 2-abelian-cubes are avoidable over a binary alphabet and that 3-abelian-squares are avoidable over a ternary alphabet, answering positively to two questions of Karhumäki et al.. We also show the existence of infinite additive-cube-free words on several ternary alphabets. To achieve this, we give sufficient conditions for a morphism to be k-abelian-n-power-free (resp. additive-n-power-free), and then we give several morphisms which respect these conditions. Additionally, all our constructions show that the number of such words grows exponentially. As a corollary, we get a new lower bound of 31/19 = 1.059526 . . . for the growth rate of abelian-cube-free words. Keywords: Combinatorics on words, k-abelian equivalence, square-free, cube-free, morphism
1. Introduction Avoidability of repetitions in words is one of the most studied topics in word combinatorics since the seminal papers of Thue [26, 27]. One famous example is Dejean’s conjecture, recently solved by several authors (see [23]). The avoidability of abelian repetitions received a lot of interest since a question from Erdös in 1957 [9, 10]. Two words u, v ∈ A∗ are abelian equivalent, denoted u ≡a v, if for every a ∈ A, |u|a = |v|a . A word u is an abelian-n-power , where n ≥ 2, if u = u1 u2 . . . un such that ui ≡a ui+1 for every i ∈ {1, . . . , n − 1}. An abelian square (resp. abelian cube) is an abelian-2-power (resp. abelian-3-power). It is not difficult to see that every ternary word of size at least 8 has an abelian square. Erdös [9, 10] raised the question whether they can be avoided in an infinite word on an alphabet of size 4. Evdokimov [11] showed that one can avoid them on an alphabet of size 25, which was later lowered to 5 by Pleasants [22]. Finally, Keränen [18] answered positively to Erdos’s question in 1992. Furthermore, Dekking [7] showed that abelian cubes can be avoided in an infinite ternary word, and that abelian-4-powers can be avoided in an infinite binary word.
Email address: [email protected] (Michaël Rao)
Preprint submitted to Elsevier
December 15, 2014
We are here interested in two variations of the previous problem. The first one is the k-abelian-equivalence introduced by Karhumäki et al. [14, 16, 17]. Let k ≥ 1. Two words u and v (u, v ∈ A∗ ) are k-abelian-equivalent, denoted u ≡a,k v, if for every w ∈ A∗ with |w| ≤ k, |u|w = |v|w . A word u is a k-abeliann-power , n ≥ 2, if u = u1 u2 . . . un such that ui ≡a,k ui+1 for every i ∈ {1, . . . , n − 1}. A k-abelian-square (resp. k-abelian-cube) is a k-abelian-2-power (resp. k-abelian-3-power). This notion is between the abelian equivalence (which is the 1-abelian-equivalence) and the usual equality between words (which can be viewed as the ∞-abelian-equivalence). Since cubes are avoidable in the binary alphabet (e.g. in the Prouhet-Thue-Morse word), but are not avoidable in the abelian sense, it is natural to ask for the smallest k for which k-abelian-cubes are avoidable on a binary alphabet. In [14] authors showed that k ≤ 8, and in [20] that k ≤ 5. Finally, in [21], Mercaş and Saarela showed that k ≤ 3. The same question can be asked for k-abelian-squares on a ternary alphabet: 2-abeliansquares cannot be avoided [15], but Huova showed that 64-abelian-squares can be avoided [12]. In Section 2, we give sufficient conditions for a morphism h : A∗ → B ∗ to be k-abelian-n-power-free (for a fixed n ≥ 2 and k ≥ 1), that is for every abelian-n-power-free word w ∈ A∗ , h(w) is k-abelian-n-power-free. Then we give morphisms which respect the conditions, in order to construct 2-abeliancube-free binary words and 3-abelian-square-free ternary words. This answers the two previous questions and also prove that the number of such words grows exponentially, as abelian-square-free on four letters [3], and abelian-cube-free ternary words ([1], see also Section 3). The second notion is the additive-cube-avoidability. A word w ∈ N∗ is an additive cube ifP w = pqr, p, q and r are non-empty-word such that P whereP |p| = |q| = |r| and (p) = (q) = (r). A word is additive-cube-free if it has no factor which is an additive cube. Clearly, such words are also abelian-cubefree. Recently Cassaigne et al. [5] showed that one can construct an infinite additive-cube-free word on the alphabet {0, 1, 3, 4}. The question of infinite additive-square-free word’s existence on a finite alphabet is still open. ∗ In Section 3 we give sufficient conditions for a substitution h : A∗ → 2B , A, B ⊆ N, to be additive-cube-free. We present substitutions from the alphabet {0, 1, 3, 4} to several ternary alphabets which respects these conditions. Moreover, the presented constructions show directly that the number of additive-cubefree words on these ternary alphabets grows exponentially. The lower bound of 31/19 = 1.059526 . . . we obtain for the growth rate for the alphabet {0, 1, 8} is also a new lower bound for the number of abelian-cube-free words on a ternary alphabet. 2. k-abelian-n-power-free morphisms 2.1. Preliminaries Let |u|w denote the number of occurrences of the factor w in u. The Parikh vector of a word u ∈ A∗ , where A = {a1 , a2 , . . . , ak }, is Ψ(u) = (|u|a1 , |u|a2 , . . . , |u|ak ). For a set S ⊆ A∗ , ΨS (u) is the vector indexed by S such that 2
ΨS (u)[w] = |u|w for every w ∈ S. When the alphabet is clear in the context, we let Ψk (u) be ΨAk (u), for k ≥ 1. Let Pref(u) be the set of prefixes of u, and Suf(u) be its set of suffixes. For k ≥ 0, let pref k (u) (resp. suf k (u)) be the prefix (resp. suffix) of u of size k. There are several equivalent definitions for k-abelian-equivalence (see [17]). Two words u and v of size at most k − 1 are k-abelian-equivalent if and only if they are equal. Otherwise, the following conditions are equivalent: • u and v are k-abelian-equivalent (i.e. u ≡a,k v). • For every w ∈ A∗ with |w| ≤ k, |u|w = |v|w . • For every w ∈ Ak , |u|w = |v|w , pref k−1 (u) = pref k−1 (v) and suf k−1 (u) = suf k−1 (v). • For every w ∈ Ak , |u|w = |v|w , and pref k−1 (u) = pref k−1 (v). Given k ≥ 1 and n ≥ 2, a (possibly infinite) word w is k-abelian-n-power-free if no non-empty factor in w is a k-abelian-n-power. A word is k-abelian-squarefree (resp. k-abelian-cube-free) if it is k-abelian-2-power-free (resp. k-abelian-3power-free). A morphism h : A∗ → B ∗ is k-abelian-n-power-free if for every abelian-npower-free word u ∈ A∗ , h(u) is k-abelian-n-power-free. Note that u has to be abelian-n-power-free, not only k-abelian-n-power-free; we explain in Section 2.4 why we use this weaker notion. A morphism h : A∗ → B ∗ is k-abelian-squarefree (resp. k-abelian-cube-free) if it is k-abelian-2-power-free (resp. k-abelian-3power-free). 2.2. Testing k-abelian-n-power-freeness In [2], Carpi gave a set of conditions which assures that a given morphism is abelian-n-power-free. We give in the following theorem a set of similar conditions which assures that a given morphism is k-abelian-n-power-free. Theorem 1. We fix k ≥ 1 and n ≥ 2, and two alphabets A and B. Let h : A∗ → B ∗ be a morphism. Suppose that: (i) For every abelian-n-power-free word w ∈ A∗ with |w| ≤ 2 or |h(w[2 : |w| − 1])| ≤ (k − 2)n − 2, h(w) is k-abelian-n-power-free. (ii) There are p, s ∈ B k−1 such that for every a ∈ A, p = pref k−1 (h(a)p) and s = suf k−1 (sh(a)). (iii) The matrix N indexed by B k × A, with N [w, x] = |h(x)p|w , has rank |A|. (iv) Let S ⊆ B k , with |S| = |A|, such that the matrix M indexed by S × A, with M [w, x] = |h(x)p|w , is invertible. Let ΨS (v, u) = ΨS (vp) + ΨS (su) − ΨS (sp) and Ψk (v, u) = ΨB k (v, u). For every ai ∈ A and ui , vi ∈ A∗ with ui vi = h(ai ); 0 ≤ i ≤ n; such that: 3
(P) |{pref k−1 (vi p) : 0 ≤ i < n}| = 1, (I) M −1 (ΨS (vi−1 , ui ) − ΨS (vi , ui+1 )) is an integer vector, for every 1 ≤ i < n, (C) Ψk (vi−1 , ui ) − Ψk (vi , ui+1 ) ∈ im(N ) for every 1 ≤ i < n, there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that for every 1 ≤ i < n : M −1 ΨS (vi−1 , ui ) − (1 − αi−1 )Ψ(ai−1 ) − αi Ψ(ai ) = M −1 ΨS (vi , ui+1 ) − (1 − αi )Ψ(ai ) − αi+1 Ψ(ai+1 ). (1) Then h is k-abelian-n-power-free. Proof. Suppose that h(w) has a k-abelian-n-power q1 . . . qn . Let q0 and qn+1 be such that h(w) = q0 q1 . . . qn qn+1 . By condition (i), if |q1 | < k − 1, then w has an abelian-n-power. So we have |qi | ≥ k − 1 for every 1 ≤ i ≤ n. There are, for every 0 ≤ i ≤ n, ai ∈ A, ui ∈ Pref(h(ai )) and ri ∈ A∗ such that, for every 0 ≤ i ≤ n, r0 . . . ri ai ∈ Pref(w) and q0 . . . qi = h(r0 . . . ri )ui . Note that, for a 1 ≤ i ≤ n, ri can be empty, but ai is always the first letter of ri+1 ai+1 . Let vi be such that ui vi = h(ai ) for every 0 ≤ i ≤ n. By condition (i), one can suppose w.l.o.g. that |r1 . . . rn an | ≥ 3. By condition (ii), for every 1 ≤ i ≤ n, pref k−1 (qi ) = pref k−1 (vi−1 p). Since q1 . . . qn is a k-abelian-n-power, we have condition (P). Claim 1. Let r ∈ A∗ and u, v ∈ B ∗ . Then: • N Ψ(r) = Ψk (h(r)p) = Ψk (sh(r)) = Ψk (sh(r)p) − Ψk (sp), • Ψk (vh(r)p) = Ψk (vp) + N Ψ(r), • Ψk (sh(r)u) = Ψk (su) + N Ψ(r). Proof. If pref k−1 (u) = p, then Ψk (vu) = Ψk (vp) + Ψk (u). Similarly, if suf k−1 (v) = s, then Ψk (vu) = Ψk (v) + Ψk (su). All the equalities follow from the previous facts, and the definition of N . Claim 2. For every 1 ≤ i ≤ n: Ψk (qi ) = N (Ψ(ri ) − Ψ(ai−1 )) + Ψk (vi−1 , ui ).
(2)
Proof. By double counting, we have : Ψk (qi ) + Ψk (sh(ri ai )p) = Ψk (sh(ri )ui ) + Ψk (vi−1 h(a−1 i−1 ri ai )p). By Claim 1: Ψk (qi ) + N Ψ(ri ai ) + Ψk (sp) = Ψk (sui ) + N Ψ(ri ) + Ψk (vi−1 p) + N Ψ(a−1 i−1 ri ai ). Thus: Ψk (qi ) = Ψk (vi−1 , ui ) + N (Ψ(ri ) − Ψ(ai−1 )). 4
Since Ψk (qi ) = Ψk (qi+1 ) for every 1 ≤ i < n, we have the condition (C). Now we have directly ΨS (qi ) = M (Ψ(ri ) − Ψ(ai−1 )) + ΨS (vi−1 , ui ). Since ΨS (qi ) = ΨS (qi+1 ): M −1 (ΨS (vi−1 , ui ) − ΨS (vi , ui+1 )) = Ψ(ri+1 ) − Ψ(ai ) − Ψ(ri ) + Ψ(ai−1 ). The right part is an integer vector, so we have condition (I). Thus, by condition (iv), there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that (1) is fulfilled. Equation (1) together with equation (2) give: − Ψ(ri ) + Ψ(ai−1 ) − (1 − αi−1 )Ψ(ai−1 ) − αi Ψ(ai ) = −Ψ(ri+1 ) + Ψ(ai ) − (1 − αi )Ψ(ai ) − αi+1 Ψ(ai+1 ) that is: Ψ(ri ) − αi−1 Ψ(ai−1 ) + αi Ψ(ai ) = Ψ(ri+1 ) − αi Ψ(ai ) + αi+1 Ψ(ai+1 ).
(3)
In equation (3), either the left or the right part is a non-negative vector. Since equation (3) is fulfilled for every 1 ≤ i < n, Ψ(ri ) − αi−1 Ψ(ai−1 ) + αi Ψ(ai ) is −α i a non negative vector for every 1 ≤ i ≤ n. Let ri0 = ai−1i−1 ri aα i ; 1 ≤ i ≤ n. 0 Since ai is the first letter of ri ai+1 , and Ψ(ri ) = Ψ(ri ) − αi−1 Ψ(ai−1 ) + αi Ψ(ai ) is a non-negative vector, ri0 is well defined in B ∗ . In one hand r10 . . . rn0 is a factor of w, and is non empty since |r10 . . . rn0 | ≥ |r1 . . . rn an | − 2. On the other 0 hand Ψ(ri0 ) = Ψ(ri+1 ) (by equation 3), for every 1 ≤ i < n. Thus, w has an abelian-n-power r10 . . . rn0 . We introduce ΨS (v, u) in order to handle pairs (v, u) such that |vu| < k − 1 (otherwise we have Ψk (v, u) = Ψk (vu)). Theorem 1 gives a set of sufficient conditions, but are still far from a characterization, as Carpi partially done for abelian-n-power-free morphisms [2]. The key point is the condition (ii). One mentions that we can save up the suffix condition in (ii) by carefully handling the cases where ui or vi has size less than k. However, we still need either the prefix (or the suffix) condition in order to properly define N . 2.3. 2-abelian-cube-free and 3-abelian-square-free morphisms Morphisms h2 and h02 respect the conditions of Theorem 1 for k = 2 and n = 3, i.e. are 2-abelian-cube-free, while morphisms h3 and h03 respect the conditions for k = 3 and n = 2, i.e. are 3-abelian-square-free. The checks were done by computer, and took only a few seconds. Thus, the infinite word h2 (u) (resp. h02 (u)) where u is an infinite abelian-cube-free word (for example a fixed point of Dekking’s morphism µ : 0 → 0012, 1 → 112, 2 → 022 [7]) is a 2-abelian-cube-free binary word. Similarly, h3 (v) (resp. h03 (v)), where v is an infinite abelian-square-free word on an alphabet of size 4 (for example, a fixed point of Keränen’s morphism g85 [18]), is an infinite 3-abelian-square-free ternary word. Over all the 2-abelian-cube-free morphisms we found, h2 is the smallest uniform morphism, while h02 is the one which minimize |h(012)|. If we are only 5
2-abelian-cube-free morphisms: 0 → 00100101001011001001010010011001001100101101011 h2 : 1 → 00100110010011001101100110110010011001101101011 2 → 00110110101101001011010110100101001001101101011 0 → 00100101001100100101001001100100110011011 0 h2 : 1 → 010110110011011001100100110011011 2 → 0101101001010010110011011 3-abelian-square-free morphisms: 0 → 0102012021012010201210212 1 → 0102101201021201210120212 h3 : 2 → 0102101210212021020120212 3 → 0121020120210201210120212 0 → 01201020120212012101201021 1 → 01202120121021201021 h03 : 2 → 0120210201021 3 → 0121020121 Morphisms such that h(µ∞ (0)) is 2-abelian-cube-free: 0 → 001001100110110011001001100100101 hd : 1 → 001011010110100101001001100100101 2 → 001011010110110011001001101011011 0 → 0101101001011 0 hd : 1 → 010110110011011001100100110011011 2 → 00100101001001100100110011011 Table 1: Morphisms for k-abelian-n-power-free words.
6
interested in 2-abelian-cube-free infinite word, one can find simpler construction. The morphism hd ◦ µ is 2-abelian-cube-free so hd (µ∞ (0)) is 2-abelian-cube-free. We also claim that h0d (µ∞ (0)) is 2-abelian-cube-free. One can modify the decision procedure of Theorem 1 to compute the set of “patterns” that u has to avoid to ensure that h(u) is k-abelian-n-power-free. This notion of patterns was used by Carpi [3, 4] to prove that a substitution is abelian-square free, or by Keränen [19] to prove that a fixed point of g98 is abelian-square free, even though g98 is not abelian-square free. This was also used, under the name of template, by Aberkane et al. [1] to show the exponential growth rate of abelian-cube-free ternary words, and by Currie and Rampersad [6] for an algorithm which decide if a fixed point of a morphism is abelian-n-power-free. More recently, Mercaş and Saarela [20, 21] used this kind of patterns to show that a morphic word is k-abelian-cube-free. Doing this, we are able to show that h0d ◦ µ3 (u) is 2-abelian-cube free if and only if u forbids factors of the form F = {pqr, 1p0q0r2 : Ψ(p) = Ψ(q) = Ψ(r)} ∪ {0p1q0r2, 1p1q0r2 : Ψ(p1) = Ψ(q0) = Ψ(r0)}. Moreover, µ(u) forbids factors of the form F if and only if u forbids factors of the form F (in other words, µ is F -free). Thus, h0d (µ∞ (0)) is 2-abelian-cube-free, but for every n ≥ 0, h0d ◦ µn is not 2-abelian-cube-free (e.g. for every n ≥ 0, h0d (µn (1002)) has a 2abelian-cube). 2.4. Final remarks and questions We finally shortly explain why we use this weak notion of k-abelian-n-powerfreeness for morphisms. On one hand, k-abelian-squares cannot be avoided by a pure morphic word on a ternary alphabet [13]. So there is no morphism h : {0, 1, 2} → {0, 1, 2} such that for every k-abelian-square-free word u, h(u) is k-abelian-square-free, except trivial ones. On the other hand, suppose that there is a morphism h : A∗ → B ∗ , with |A| > |B|, such that for every 2-abeliancube-free word u ∈ A∗ , h(u) is 2-abelian-cube-free. Without lost of generality, there is {a, b} ⊆ A, such that the first letter of h(a) and h(b) is the same. Then babbababb is 2-abelian-cube-free, but h(bab) ≡a,2 h(abb) thus h(babbababb) is an 2-abelian-cube. We have a contradiction, so such a morphism cannot exist. Nevertheless, we cannot conclude directly when |A| = |B| and the first and last letters of the images differ. More specifically, the following question is still open. Question 1. Is there a pure morphic binary word which avoids 2-abelian-cubes ? Let us also raise some questions on avoidability of long repetitions. Every infinite binary word contains arbitrarily long abelian squares, while ones exist which avoid squares of period at least 3 [8, 25]. We recently showed that one can avoid 3-abelian-squares of period at least 3 over a binary alphabet [24]. It seems natural to ask the following: Question 2. Is there a p ∈ N such that 2-abelian-squares of period at least p can be avoided over a binary alphabet ? 7
That reminds the questions suggested by Mäkelä (see [19]): Question 3. (1) Can we avoid abelian-squares of the form uv, with |u| ≥ 2, over a ternary alphabet ? (2) Can we avoid abelian-cubes of the form uvw, with |u| ≥ 2, over a binary alphabet ? In [24], we answer negatively to (1). Then we modify this question to the following one: Question 4. Is there a p ∈ N such that one can avoid abelian cubes of period at least p over a binary alphabet ? 3. Ternary words avoiding additive cubes 3.1. Testing additive-n-power-freeness Let Σ be the morphism from the free monoid on the alphabet N to the additive group (Z, +) such that Σ(x) = x for every x ∈ N. A word w ∈ N∗ is an additive-n-power , with n ≥ 2, if w = p1 . . . pn , such that for every 1 ≤ i < n, |pi | = |pi+1 | and Σ(pi ) = Σ(pi+1 ). A word is an additive-cube (resp. additivesquare) if it is an additive-3-power (additive-2-power). A (possibly infinite) word w is additive-n-power-free if no non-empty factor of w is an additive-n-power. Clearly, such words are also abelian-n-power-free. In [5], authors prove that the fixed point of the morphism 0 → 03, 1 → 43, 3 → 1, 4 → 01 is additive-cube-free. ∗ A substitution is a morphism s : A∗ → 2B between the free monoid A∗ and the power monoid of B ∗ , that is the monoid of subsets of B ∗ , with the operation U · V = {uv : (u, v) ∈ U × V }. A morphism h : A∗ → B ∗ can be ∗ viewed as a substitution s : A∗ → 2B such that s(w) = {h(w)}. A substitution ∗ s : A∗ → 2B , where A, B ⊆ N, is additive-n-power-free if for every additive-npower-free word u ∈ A∗ , every v ∈ s(u) is additive-n-power-free. We give sufficient conditions for a substitution to be additive-n-power-free in the following theorem. ∗
Theorem 2. We fix n ≥ 2 and A, B ⊆ N. Let s : A∗ → 2B be a substitution. Suppose that: (i) For every additive-n-power-free word w0 ∈ A∗ with |w0 | ≤ 2, every w ∈ s(w0 ) is additive-n-power-free. (ii) There is (l, γ, β) ∈ N × Z × Z, with β 6= 0, such that for every a ∈ A and w ∈ s(a), we have |w| = l and Σ(w) = γ + aβ. (iii) For every ai ∈ A, wi ∈ s(ai ), and ui , vi ∈ A∗ with ui vi = wi ; 0 ≤ i ≤ n; such that for every 1 ≤ i < n: (L) |vi−1 ui | ≡ |vi ui+1 | (mod l), 8
s015
s016
s017
s027
s037
s018
s038
s019
s029
s049
0 → {005015100100115010115, 005015100100115100115} 1 → {005015100100105055115, 050015100100105055115} : 3 → {005015101155155055115, 050015101155155055115} 4 → {005015155055155055115, 050015155055155055115} 0 → {00101160101006001016, 00101160101006001106} 1 → {00166060101006001016, 00166060101006001106} : 3 → {00166166110160661106, 00166166110166061106} 4 → {00166166066160661106, 00166166066166061106} 0 → {00170010011711001071, 00170010011711001701} 1 → {00170017707001001071, 00170017707001001701} : 3 → {00170017711017177077, 01070017711017177077} 4 → {00170017707077177077, 01070017707077177077} 0 → {0020720220220722007, 0020720220227022007} 1 → {7220720220220722007, 7220720220227022007} : 3 → {7077200770720722007, 7077200770727022007} 4 → {7077272770720722007, 7077272770727022007} 0 → {00300307303037707307, 00300307303037700737, 00300307303037707037} 1 → {00300300707737700737, 00300300707737707037, 00300300707737707307} : 3 → {00337730337737700737, 00337730337737707037, 00337730337737707307} 4 → {00337737707737700737, 00337737707737707307, 00337737707737707037} 0 → {0081001008011811011, 0081010080011811011, 0081001080011811011} 1 → {0081001008011818008, 0081010080011818008, 0081001080011818008} : 3 → {0081018818808811811, 0081108818808811811, 0081810818808811811} 4 → {0081018818808808188, 0081108818808808188, 0081188018808808188} 0 → {003800303830033833003, 003800308330033833003} 1 → {003800303830080038388, 003800308330080083838} : 3 → {003808833833038838838, 003808838330338838838} 4 → {003808838388088388388, 083008838388088388388} 0 → {0090110191001009, 0090110911001009} 1 → {0090119110110199, 0900119110110199} : 3 → {0090190090099199, 0900190090099199} 4 → {0090119199099199, 0900119199099199} 0 → {00290020020090022029, 00290020020090020229} 1 → {00290099220090022029, 00290099220090020229} : 3 → {00220292299099299099, 22920220099099299099} 4 → {22920992299099299099, 22990292299099299099} 0 → {00400400900499009, 00400400900949009} 1 → {00400449440099409, 00400449440499009} : 3 → {00409909909499409, 00409909949099409} 4 → {44944909949499009, 44944909949909409} Table 2: Additive-cube-free substitutions.
9
(M) Σ(vi−1 ui ) ≡ Σ(vi ui+1 ) + xi γ (mod β), (where xi = (|vi−1 ui | − |vi ui+1 |)/l for every 1 ≤ i < n) there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that for every 1 ≤ i < n: (a) αi − αi−1 = xi + αi+1 − αi , (b) Σ(vi−1 ui ) + β[(αi−1 − 1)ai−1 − αi ai ] = Σ(vi ui+1 ) + γxi + β[(αi − 1)ai − αi+1 ai+1 ]. Then s is additive-n-power-free. Proof. Suppose that w ∈ s(w0 ) has an additive-n-power q1 . . . qn . Let q0 and qn+1 be such that w = q0 q1 . . . qn qn+1 . For every 0 ≤ i ≤ n, there is ai ∈ A, wi ∈ s(ai ), ui ∈ Pref(wi ), and ri ∈ A∗ such that r0 . . . ri ai ∈ Pref(w0 ) and q0 . . . qi ∈ s(r0 . . . ri ) · {ui }. Let vi be such that ui vi = wi for every 0 ≤ i ≤ n. By condition (i), one can suppose w.l.o.g. that |r1 . . . rn an | ≥ 3. By condition (ii), for every p ∈ s(p0 ), we have Σ(p) = γ|p0 | + βΣ(p0 ). For every 1 ≤ i ≤ n, we have ui−1 qi ∈ s(ri ) · {ui }. Thus, by condition (ii), and by the fact that ui−1 vi−1 = wi−1 we have: |qi | = |vi−1 ui | + l(|ri | − 1)
(4)
Σ(qi ) = γ(|ri | − 1) + β(Σ(ri ) − ai−1 ) + Σ(vi−1 ui ).
(5)
and By equation (4) and by the fact that for every 1 ≤ i < n, |qi | = |qi−1 |, we have the condition (L), and: |ri+1 | − |ri | = (|vi−1 ui | − |vi ui+1 |)/l = xi . Since for every 1 ≤ i < n, Σ(qi ) = Σ(qi−1 ), we have: γ(|ri | − 1) + β(Σ(ri ) − ai−1 ) + Σ(vi−1 ui ) = γ(|ri+1 | − 1) + β(Σ(ri+1 ) − ai ) + Σ(vi ui+1 ). (6) Thus Σ(vi−1 ui ) = Σ(vi ui+1 ) + γxi + β(Σ(ri+1 ) − ai − Σ(ri ) + ai−1 ), and equation (M) is fulfilled. So, by condition (iii), there is (α0 , . . . , αn ) ∈ {0, 1}n+1 such that (a) and (b) are fulfilled. By equation (a), we have, for every 1 ≤ i < n; |ri | + αi − αi−1 = |ri−1 | + αi+1 − αi .
(7)
If ri is empty, ai = ai+1 otherwise the first letter of ri is ai . In equation (7), the right side or the left side must be non-negative. Thus, for every 1 ≤ i ≤ n, 10
−α
i |ri | + αi − αi−1 ≥ 0, and ri0 = ai−1i−1 ri aα i ; 1 ≤ i ≤ n; is well defined. We have 0 0 |ri | = |ri | + αi − αi−1 and Σ(ri ) = Σ(ri ) + αi ai − αi−1 ai−1 . By equation (7), 0 for every 1 ≤ i < n, |ri0 | = |ri+1 |. Moreover r10 . . . rn0 is a factor of w0 , and is non 0 0 empty since |r1 . . . rn | ≥ |r1 . . . rn an | − 2. 0 ). Thus, Σ(ri0 ) = When we subtract (b) to (6), we get βΣ(ri0 ) = βΣ(ri+1 0 0 Σ(ri+1 ) for every 1 ≤ i < n, and w has an additive-n-power r10 . . . rn0 .
Theorem 2 can be used to find additive-square-free, additive-cube-free and additive-4-power-free substitutions. However, we have few hopes to find an additive-square-free substitution, while additive-4-powers are equivalent to abelian-4-powers on binary words. 3.2. Additive-cube-free substitutions We have checked by computer that every substitution in Table 2 respects the conditions of Theorem 2. Since there is an infinite additive-cube-free word on the alphabet {0, 1, 3, 4}, one can construct infinite additive-cube-free words on the alphabets {0, 1, 5}, {0, 1, 6}, {0, 1, 7}, {0, 2, 7}, {0, 3, 7}, {0, 1, 8}, {0, 3, 8}, {0, 1, 9}, {0, 2, 9} and {0, 4, 9}. In our substitutions, each letter has at least two images. This clearly shows that number of additive-cube-free words on these alphabets grows exponentially. For the alphabet {0, 1, 8}, we got 3 images of size 19 for each letter, giving the lower bound of 31/19 = 1.059526 . . . for the growth rate. This bound is also a new lower bound for the growth rate of abelian-cube-free words on ternary alphabet. (The previous known bound was 21/24 = 1.029302 . . . in [1].) We conjecture that for every alphabet A = {0, i, j} such that i and j are coprime and j ≥ 6, there exists an infinite additive-cube-free word on the alphabet A. The cases {0, 1, 2}, {0, 1, 3}, {0, 1, 4} and {0, 2, 5} are left open. Furthermore, it seems difficult to construct a very long word on the alphabet {0, 1, 2, 3} avoiding additive cubes (the longest we got has size ∼ 1.4 × 105 ). Question 5. Are there infinite additive-cube-free words on the following alphabets : {0, 1, 2, 3}, {0, 1, 4} and {0, 2, 5} ? The substitutions in Table 3 also respect the conditions of Theorem 2, thus the existence of an infinite additive-cube-free word on the alphabet {0, 1, 2, 3} imply the existence of infinite additive-cube-free words on alphabets {0, 1, 4} and {0, 2, 5}. Acknowledgments The author would like to thank Pascal Ochem for valuable discussions on these problems. [1] A. Aberkane, J. D. Currie, N. Rampersad. The number of ternary words avoiding abelian cubes grows exponentially. J. Integer Sequences, Vol. 7.2, Art. 04.2.7 (2004).
11
s014
s025
0 → {004114104011011004011} 1 → {004114104011011014144} : 2 → {004114104010044044144} 3 → {004114104044144044144} 0 → {02200520220250552} 1 → {02252520220250552} : 2 → {02255055200550552} 3 → {02255055252550552}
Table 3: Additive-cube-free substitutions from {0, 1, 2, 3}∗ .
[2] A. Carpi. On Abelian Power-Free Morphisms. Int. J. Algebra Comput., Vol. 03(2) 151–167 (1993). [3] A. Carpi. On the number of Abelian square-free words on four letters. Disc. Appl. Math, Vol. 81(1-3), 155–167 (1998). [4] A. Carpi. On Abelian squares and substitutions. Theoret. Comput. Sci., Vol. 218(1), 61–81 (1999). [5] J. Cassaigne, J. D. Currie, L. Schaeffer, J. Shallit. Avoiding three consecutive blocks of the same size and same sum. J. ACM, Vol. 61(2), Art. 10 (2014). [6] J. D. Currie, N. Rampersad. Fixed points avoiding Abelian k-powers. J. Comb. Theory Ser. A, Vol. 119(5) 942–948 (2012). [7] F.M. Dekking. Strongly nonrepetitive sequences and progression-free sets. J. Comb. Theory Ser. A, Vol. 27(2), 181–185 (1979). [8] R.C Entringer, D.E Jackson, J.A Schatz. On nonrepetitive sequences. J. Comb. Theory Ser. A, Vol. 16(2), 159–164 (1974). [9] P. Erdös. Some unsolved problems. Michigan Math. J., Vol. 4(3), 291–300 (1957). [10] P. Erdös. Some unsolved problems. Magyar Tud. Akad. Mat. Kutató Int. Közl, Vol. 6, 221–254 (1961). [11] A. A. Evdokimov. Strongly asymmetric sequences generated by a finite number of symbols. Dokl. Akad. Nauk. SSSR, Vol. 179, 1268–1271 (1968); Soviet Math. Dokl., Vol. 9, 536–539 (1968). [12] M. Huova. Existence of an infinite ternary 64-abelian square-free word. RAIRO - Theor. Inf. and Applic., Vol. 48(3), 307–314 (2014).
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[13] M. Huova, J. Karhumäki. On the Unavoidability of k-Abelian Squares in Pure Morphic Words. J. Integer Sequences, Vol. 16, Art. 13.2.9 (2013). [14] M. Huova, J. Karhumäki, A. Saarela. Problems in between words and abelian words: k-abelian avoidability. Theoret. Comput. Sci., Vol. 454, 172–177 (2012). [15] M. Huova, J. Karhumäki, A. Saarela, K. Saari. Local squares, periodicity and finite automata. Rainbow of Computer Science, LNCS Vol. 6570, 90– 101 (2011). [16] J. Karhumäki. Generalized Parikh mappings and homomorphisms. Inform. Control, Vol. 47, 155–165 (1980). [17] J. Karhumäki, A. Saarela, L. Zamboni. On a generalization of Abelian equivalence and complexity of infinite words. J. Comb. Theory Ser. A, Vol. 120(8), 2189–2206 (2013). [18] V. Keränen. Abelian squares are avoidable on 4 letters. Proc. of ICALP 1992, LNCS, Vol. 623, 41–52 (1992). [19] V. Keränen. New Abelian Square-Free DT0L-Languages over 4 Letters. Manuscript (2003). [20] R. Mercaş, A. Saarela. 5-abelian cubes are avoidable on binary alphabets. Proc. of the 14th Mons Days of Theoret. Comput. Sci. (2012). [21] R. Mercaş, A. Saarela. 3-Abelian Cubes Are Avoidable on Binary Alphabets. Proc. of DLT 2013, LNCS Vol. 7907, 374–383 (2013). [22] P. A. B. Pleasants. Non-repetitive sequences. Proc. Cambridge Philos. Soc., Vol. 68, 267–274 (1970). [23] M. Rao. Last cases of Dejean’s conjecture. Theoret. Comput. Sci., Vol. 412(27), 3010–3018 (2011). [24] M. Rao, M. Rosenfeld. Avoidability of long k-abelian repetitions. Proc. of the 15th Mons Days of Theoret. Comput. Sci. (2014). [25] A. S. Fraenkel, R. J. Simpson. How many squares must a binary sequence contain? Electron. J. Combin., Vol. 2, (1995). [26] A. Thue. Über unendliche Zeichenreihen. Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, Vol. 7, 1–22 (1906). [27] A. Thue. Über die gegenseitige Lage gleicher Teile gewisser Zeichenreihen. Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, Vol. 10, 1–67 (1912).
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